Enter a problem...
Calculus Examples
x3ex
Step 1
Step 1.1
Differentiate using the Product Rule which states that ddx[f(x)g(x)] is f(x)ddx[g(x)]+g(x)ddx[f(x)] where f(x)=x3 and g(x)=ex.
x3ddx[ex]+exddx[x3]
Step 1.2
Differentiate using the Exponential Rule which states that ddx[ax] is axln(a) where a=e.
x3ex+exddx[x3]
Step 1.3
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
x3ex+ex(3x2)
Step 1.4
Simplify.
Step 1.4.1
Reorder terms.
exx3+3exx2
Step 1.4.2
Reorder factors in exx3+3exx2.
x3ex+3x2ex
x3ex+3x2ex
x3ex+3x2ex
Step 2
Step 2.1
By the Sum Rule, the derivative of x3ex+3x2ex with respect to x is ddx[x3ex]+ddx[3x2ex].
f′′(x)=ddx(x3ex)+ddx(3x2ex)
Step 2.2
Evaluate ddx[x3ex].
Step 2.2.1
Differentiate using the Product Rule which states that ddx[f(x)g(x)] is f(x)ddx[g(x)]+g(x)ddx[f(x)] where f(x)=x3 and g(x)=ex.
f′′(x)=x3ddx(ex)+exddx(x3)+ddx(3x2ex)
Step 2.2.2
Differentiate using the Exponential Rule which states that ddx[ax] is axln(a) where a=e.
f′′(x)=x3ex+exddx(x3)+ddx(3x2ex)
Step 2.2.3
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
f′′(x)=x3ex+ex(3x2)+ddx(3x2ex)
f′′(x)=x3ex+ex(3x2)+ddx(3x2ex)
Step 2.3
Evaluate ddx[3x2ex].
Step 2.3.1
Since 3 is constant with respect to x, the derivative of 3x2ex with respect to x is 3ddx[x2ex].
f′′(x)=x3ex+ex(3x2)+3ddx(x2ex)
Step 2.3.2
Differentiate using the Product Rule which states that ddx[f(x)g(x)] is f(x)ddx[g(x)]+g(x)ddx[f(x)] where f(x)=x2 and g(x)=ex.
f′′(x)=x3ex+ex(3x2)+3(x2ddx(ex)+exddx(x2))
Step 2.3.3
Differentiate using the Exponential Rule which states that ddx[ax] is axln(a) where a=e.
f′′(x)=x3ex+ex(3x2)+3(x2ex+exddx(x2))
Step 2.3.4
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
f′′(x)=x3ex+ex(3x2)+3(x2ex+ex(2x))
f′′(x)=x3ex+ex(3x2)+3(x2ex+ex(2x))
Step 2.4
Simplify.
Step 2.4.1
Apply the distributive property.
f′′(x)=x3ex+ex(3x2)+3(x2ex)+3(ex(2x))
Step 2.4.2
Combine terms.
Step 2.4.2.1
Multiply 2 by 3.
f′′(x)=x3ex+ex(3x2)+3x2ex+6(ex(x))
Step 2.4.2.2
Add ex(3x2) and 3x2ex.
Step 2.4.2.2.1
Move ex.
f′′(x)=x3ex+3x2ex+3x2ex+6exx
Step 2.4.2.2.2
Add 3x2ex and 3x2ex.
f′′(x)=x3ex+6x2ex+6exx
f′′(x)=x3ex+6x2ex+6exx
f′′(x)=x3ex+6x2ex+6exx
Step 2.4.3
Reorder terms.
f′′(x)=exx3+6exx2+6exx
Step 2.4.4
Reorder factors in exx3+6exx2+6exx.
f′′(x)=x3ex+6x2ex+6xex
f′′(x)=x3ex+6x2ex+6xex
f′′(x)=x3ex+6x2ex+6xex
Step 3
To find the local maximum and minimum values of the function, set the derivative equal to 0 and solve.
x3ex+3x2ex=0
Step 4
Step 4.1
Find the first derivative.
Step 4.1.1
Differentiate using the Product Rule which states that ddx[f(x)g(x)] is f(x)ddx[g(x)]+g(x)ddx[f(x)] where f(x)=x3 and g(x)=ex.
x3ddx[ex]+exddx[x3]
Step 4.1.2
Differentiate using the Exponential Rule which states that ddx[ax] is axln(a) where a=e.
x3ex+exddx[x3]
Step 4.1.3
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
x3ex+ex(3x2)
Step 4.1.4
Simplify.
Step 4.1.4.1
Reorder terms.
exx3+3exx2
Step 4.1.4.2
Reorder factors in exx3+3exx2.
f′(x)=x3ex+3x2ex
f′(x)=x3ex+3x2ex
f′(x)=x3ex+3x2ex
Step 4.2
The first derivative of f(x) with respect to x is x3ex+3x2ex.
x3ex+3x2ex
x3ex+3x2ex
Step 5
Step 5.1
Set the first derivative equal to 0.
x3ex+3x2ex=0
Step 5.2
Factor x2ex out of x3ex+3x2ex.
Step 5.2.1
Factor x2ex out of x3ex.
x2ex(x)+3x2ex=0
Step 5.2.2
Factor x2ex out of 3x2ex.
x2ex(x)+x2ex(3)=0
Step 5.2.3
Factor x2ex out of x2ex(x)+x2ex(3).
x2ex(x+3)=0
x2ex(x+3)=0
Step 5.3
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
x2=0
ex=0
x+3=0
Step 5.4
Set x2 equal to 0 and solve for x.
Step 5.4.1
Set x2 equal to 0.
x2=0
Step 5.4.2
Solve x2=0 for x.
Step 5.4.2.1
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
x=±√0
Step 5.4.2.2
Simplify ±√0.
Step 5.4.2.2.1
Rewrite 0 as 02.
x=±√02
Step 5.4.2.2.2
Pull terms out from under the radical, assuming positive real numbers.
x=±0
Step 5.4.2.2.3
Plus or minus 0 is 0.
x=0
x=0
x=0
x=0
Step 5.5
Set ex equal to 0 and solve for x.
Step 5.5.1
Set ex equal to 0.
ex=0
Step 5.5.2
Solve ex=0 for x.
Step 5.5.2.1
Take the natural logarithm of both sides of the equation to remove the variable from the exponent.
ln(ex)=ln(0)
Step 5.5.2.2
The equation cannot be solved because ln(0) is undefined.
Undefined
Step 5.5.2.3
There is no solution for ex=0
No solution
No solution
No solution
Step 5.6
Set x+3 equal to 0 and solve for x.
Step 5.6.1
Set x+3 equal to 0.
x+3=0
Step 5.6.2
Subtract 3 from both sides of the equation.
x=-3
x=-3
Step 5.7
The final solution is all the values that make x2ex(x+3)=0 true.
x=0,-3
x=0,-3
Step 6
Step 6.1
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
Step 7
Critical points to evaluate.
x=0,-3
Step 8
Evaluate the second derivative at x=0. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
(0)3e0+6(0)2e0+6(0)e0
Step 9
Step 9.1
Simplify each term.
Step 9.1.1
Raising 0 to any positive power yields 0.
0e0+6(0)2e0+6(0)e0
Step 9.1.2
Anything raised to 0 is 1.
0⋅1+6(0)2e0+6(0)e0
Step 9.1.3
Multiply 0 by 1.
0+6(0)2e0+6(0)e0
Step 9.1.4
Raising 0 to any positive power yields 0.
0+6⋅0e0+6(0)e0
Step 9.1.5
Multiply 6 by 0.
0+0e0+6(0)e0
Step 9.1.6
Anything raised to 0 is 1.
0+0⋅1+6(0)e0
Step 9.1.7
Multiply 0 by 1.
0+0+6(0)e0
Step 9.1.8
Multiply 6 by 0.
0+0+0e0
Step 9.1.9
Anything raised to 0 is 1.
0+0+0⋅1
Step 9.1.10
Multiply 0 by 1.
0+0+0
0+0+0
Step 9.2
Simplify by adding numbers.
Step 9.2.1
Add 0 and 0.
0+0
Step 9.2.2
Add 0 and 0.
0
0
0
Step 10
Step 10.1
Split (-∞,∞) into separate intervals around the x values that make the first derivative 0 or undefined.
(-∞,-3)∪(-3,0)∪(0,∞)
Step 10.2
Substitute any number, such as -6, from the interval (-∞,-3) in the first derivative x3ex+3x2ex to check if the result is negative or positive.
Step 10.2.1
Replace the variable x with -6 in the expression.
f′(-6)=(-6)3e-6+3(-6)2e-6
Step 10.2.2
Simplify the result.
Step 10.2.2.1
Simplify each term.
Step 10.2.2.1.1
Raise -6 to the power of 3.
f′(-6)=-216e-6+3(-6)2e-6
Step 10.2.2.1.2
Rewrite the expression using the negative exponent rule b-n=1bn.
f′(-6)=-2161e6+3(-6)2e-6
Step 10.2.2.1.3
Combine -216 and 1e6.
f′(-6)=-216e6+3(-6)2e-6
Step 10.2.2.1.4
Move the negative in front of the fraction.
f′(-6)=-216e6+3(-6)2e-6
Step 10.2.2.1.5
Raise -6 to the power of 2.
f′(-6)=-216e6+3⋅(36e-6)
Step 10.2.2.1.6
Multiply 3 by 36.
f′(-6)=-216e6+108e-6
Step 10.2.2.1.7
Rewrite the expression using the negative exponent rule b-n=1bn.
f′(-6)=-216e6+108(1e6)
Step 10.2.2.1.8
Combine 108 and 1e6.
f′(-6)=-216e6+108e6
f′(-6)=-216e6+108e6
Step 10.2.2.2
Combine fractions.
Step 10.2.2.2.1
Combine the numerators over the common denominator.
f′(-6)=-216+108e6
Step 10.2.2.2.2
Simplify the expression.
Step 10.2.2.2.2.1
Add -216 and 108.
f′(-6)=-108e6
Step 10.2.2.2.2.2
Move the negative in front of the fraction.
f′(-6)=-108e6
f′(-6)=-108e6
f′(-6)=-108e6
Step 10.2.2.3
The final answer is -108e6.
-108e6
-108e6
-108e6
Step 10.3
Substitute any number, such as -2, from the interval (-3,0) in the first derivative x3ex+3x2ex to check if the result is negative or positive.
Step 10.3.1
Replace the variable x with -2 in the expression.
f′(-2)=(-2)3e-2+3(-2)2e-2
Step 10.3.2
Simplify the result.
Step 10.3.2.1
Simplify each term.
Step 10.3.2.1.1
Raise -2 to the power of 3.
f′(-2)=-8e-2+3(-2)2e-2
Step 10.3.2.1.2
Rewrite the expression using the negative exponent rule b-n=1bn.
f′(-2)=-81e2+3(-2)2e-2
Step 10.3.2.1.3
Combine -8 and 1e2.
f′(-2)=-8e2+3(-2)2e-2
Step 10.3.2.1.4
Move the negative in front of the fraction.
f′(-2)=-8e2+3(-2)2e-2
Step 10.3.2.1.5
Raise -2 to the power of 2.
f′(-2)=-8e2+3⋅(4e-2)
Step 10.3.2.1.6
Multiply 3 by 4.
f′(-2)=-8e2+12e-2
Step 10.3.2.1.7
Rewrite the expression using the negative exponent rule b-n=1bn.
f′(-2)=-8e2+12(1e2)
Step 10.3.2.1.8
Combine 12 and 1e2.
f′(-2)=-8e2+12e2
f′(-2)=-8e2+12e2
Step 10.3.2.2
Combine fractions.
Step 10.3.2.2.1
Combine the numerators over the common denominator.
f′(-2)=-8+12e2
Step 10.3.2.2.2
Add -8 and 12.
f′(-2)=4e2
f′(-2)=4e2
Step 10.3.2.3
The final answer is 4e2.
4e2
4e2
4e2
Step 10.4
Substitute any number, such as 2, from the interval (0,∞) in the first derivative x3ex+3x2ex to check if the result is negative or positive.
Step 10.4.1
Replace the variable x with 2 in the expression.
f′(2)=(2)3e2+3(2)2e2
Step 10.4.2
Simplify the result.
Step 10.4.2.1
Simplify each term.
Step 10.4.2.1.1
Raise 2 to the power of 3.
f′(2)=8e2+3(2)2e2
Step 10.4.2.1.2
Raise 2 to the power of 2.
f′(2)=8e2+3⋅(4e2)
Step 10.4.2.1.3
Multiply 3 by 4.
f′(2)=8e2+12e2
f′(2)=8e2+12e2
Step 10.4.2.2
Add 8e2 and 12e2.
f′(2)=20e2
Step 10.4.2.3
The final answer is 20e2.
20e2
20e2
20e2
Step 10.5
Since the first derivative changed signs from negative to positive around x=-3, then x=-3 is a local minimum.
x=-3 is a local minimum
Step 10.6
Since the first derivative did not change signs around x=0, this is not a local maximum or minimum.
Not a local maximum or minimum
Step 10.7
These are the local extrema for f(x)=x3ex.
x=-3 is a local minimum
x=-3 is a local minimum
Step 11